Extension $\mathbb{Q}(\sqrt{-5}, i)/\mathbb{Q}(\sqrt{-5})$, splitting.

In the extension $\mathbb{Q}(\sqrt{-5}, i)/\mathbb{Q}(\sqrt{-5})$, must principal prime ideals of $\mathbb{Z}[\sqrt{-5}]$ necessarily split into 2? Must nonprincipal prime ideals not split?

Rephrased, does a nonzero prime ideal of $\mathbb{Z}[\sqrt{-5}]$ split completely in $\mathbb{Q}(\sqrt{-5},i)/\mathbb{Q}(\sqrt{-5})$ if and only if it is a principal ideal?

I am looking for a more elementary argument that hopefully doesn’t invoke class field theory…

Solutions Collecting From Web of "Extension $\mathbb{Q}(\sqrt{-5}, i)/\mathbb{Q}(\sqrt{-5})$, splitting."

Yes, this is true. It is a consequence of class field theory. Indeed, $H=\mathbb{Q}(\sqrt{-5},i)$ is the Hilbert class field of $K=\mathbb{Q}(\sqrt{-5})$. Thus, by class field theory, a prime of $K$ is principal if and only if it splits completely in $H/K$.

More generally, the Hilbert class field $H$ of a number field $K$ enjoys the following property:

  • Every prime ideal $\wp$ of $\mathcal{O}_K$ decomposes into the product of $h_K/f$ prime ideals in $\mathcal{O}_H$, where $f$ is the order of $[\wp]$ in the ideal class group of $\mathcal{O}_K$, and $h_K$ is the class number of $K$.

In the case of $K=\mathbb{Q}(\sqrt{-5})$, we have $h_k=2$, and $H/K$ is quadratic. If $\wp$ is principal, then $f=1$, and therefore there are $2/1=2$ primes of $H$ above $\wp$. Since $H/K$ is quadratic, that means $\wp$ is completely split. Conversely, if $\wp$ is non-principal, then $f=2$, and so there are $2/2=1$ primes of $H$ above $\wp$, i.e., $\wp$ is inert.